- 50 Views
- Uploaded on
- Presentation posted in: General

Problems in Chapter 13

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Problems in Chapter 13

Worked out for Geology/Physics 360

Sirius Now (today) in March

An Illustration of Parallax using HNSKY

Sirius in Sept. 2010 (6 months later)

Question: How would we find a closer star than Proxima Centauri?

Problem 3. The Parallax of the red giant Betelgeuse is just barely measurable and has a value of about 0.005 arc seconds. What is its distance? Suppose the measurement is in error by + or – 0.003 arc seconds. What limits can you set on its distance.

Answer: p(Betelgeuse) = 0.005 arc seconds. Therefore its distance is d = 1/p = 200 pc.

The error in this measurement is +/–0.003", so we have to add and subtract this figure from the parallax angle to give the lower and upper limits to the distance estimate.

d(pc) = 1/0.008 = 125 pc, the larger angle corresponds to the closer distance

d(pc) = 1/0.002 = 500 pc, the smaller angle corresponds to the farther distance

Problem 5. The star Rigel radiates most strongly at about 200 nm. What is its temperature? How does this compare to the Sun.

𝝀mfor Rigel = 200 nm. Use Wien’s Law to find its temperature?

T = 3 × 106 K nm / 𝝀m= 3 × 106 K nm / 200 nm = 15,000 K.

This is much hotter than the surface of the Sun, which is about 6000 K. (2.5 times hotter). (Note that we have approximated the constant here as 3 x 106, it is closer to 2.9 x 106

Problem 6. The bright southern star Alpha Centauri radiates most strongly at about 500 nm. What is its temperature? How does this compare to the Sun.

. Alpha Centauri has its peak radiation at 500 nm. Applying Wien’s Law,

T = 3 × 106 K nm / 500 nm = 6000 K.

The temperature of this star is approximately the same as our Sun’s, so Alpha Centauri has a similar spectral type to the Sun (G2).

Problem 7. Arcturus is about ½ as hot as the sun but is about 100 times more luminous. What is its radius compared to the Sun.

. For Arcturus, T = To/2, where To is the temperature of the Sun. L = 100Lo. To find the star’s radius, use the Stefan-Boltzmann Law, which relates a star’s luminosity, L, to its temperature, T, and radius, R.

L = 4𝜫R2𝝈T4 or LArc= 4𝜫R2Arc 𝝈T4Arc and Lo = 4𝜫R2o 𝝈T4o

We want to find RArc in terms of Ro.

100 Lo = LArc= 4𝜫R2Arc 𝝈T4Arc

100 (4𝜫R2o 𝝈T4o) = 4𝜫R2Arc 𝝈T4Arc

100 (R2oT4o) = R2Arc T4Arc = R2Arc (To/2) 4

100 (R2oT4o) = R2Arc (T4o) / 24

100 (R2o) = R2Arc / 16

1600 (R2o) = R2Arc

Therefore

R Arc = (1600) 1/2 Ro= 40 Ro

Arcturus is 40 times wider than the Sun.